# How is this relation not a Transitive relation?

The question says:- $A = \{1,2,3,4,5,6\}$ and $R = \{(S_1, S_2) :S_1, S_2 \subset A, S1\nsubseteq S2\}$.

My thought:- $S_1$ contains the subsets of $A$ and $S_2$ contains the subset of $A$ and $S_1$ is not a subset of $S_2$ then there are no elements common in $S_1$ and $S_2$ so there is not relations such as if $(a, b)$ is an element of $A$ and $(b, c)$ is an element of $A$ then $(a, c)$ is an element of $A$ as $b$ is not element of $S_1$.. Hence i can say the relation is transitive.. But the answer says it is not transitive

• Formatting tips here. – Em. Jul 17 '16 at 7:42
• For clarification: Let $B,C$ be two subsets of $A$. Is your relation that $B \sim C$ if $B$ is not a subset of $C$? – Christian Jul 17 '16 at 7:50

Assuming I interpreted your relation correctly, as I described in the comment...

For transitivity to hold, we require $\forall S_1, S_2, S_3$, subsets of $A$, $S_1 \sim S_2$ and $S_2 \sim S_3 \implies S_1 \sim S_3$. (or, $(S_1, S_2) \in R$ and $(S_2, S_3) \in R \implies (S_1, S_3) \in R$)

Let $S_1 = \{1, 2\}$

Let $S_2 = \{2, 3\}$

Let $S_3 = \{1, 2, 4\}$

$S_1 \sim S_2$ (i.e. $(S_1, S_2) \in R$) since $S_1$ is not a subset of $S_2$.

$S_2 \sim S_3$ (i.e. $(S_2, S_3) \in R$) since $S_2$ is not a subset of $S_3$.

But $S_1$ is a subset of $S_3$, so $S_1$ is not equivalent to $S_3$ ($(S_1, S_3) \notin R$). This is one of many examples where transitivity will fail.

• I don't know the meaning of the symbol you used in between S1 and s2 .I did some edit in the question . – danny Jul 17 '16 at 8:00
• That symbol means "Is equivalent to." It is the same thing as saying: $(S_1, S_2) \in R$. I reformatted my answer to include the notation you may be more familiar with. – Christian Jul 17 '16 at 8:01
• Thanks.. I understood where the transitivity fails. :) – danny Jul 17 '16 at 8:04